Question: Let $f$ be a polynomial function and let $f'$, its derivative, be defined as $f'(x)=x^4(x-2)(x+3)$. At how many points does the graph of $f$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
We can find the relative extrema (i.e. minima and maxima) of $f$ by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $f'(x)=x^4(x-2)(x+3)$. $f'(x)=0$ for $x=-3,0,2$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-3$, $x=0$, and $x=2$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $x< \llap{-}3$ $\llap{-}3<x<0$ $0<x<2$ $x>2$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<-3$ $x=-4$ $f'(-4)=1536>0$ $f$ is increasing $\nearrow$ $-3<x<0$ $x=-1$ $f'(-1)=-6<0$ $f$ is decreasing $\searrow$ $0<x<2$ $x=1$ $f'(1)=-4<0$ $f$ is decreasing $\searrow$ $x>2$ $x=3$ $f'(3)=486>0$ $f$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $-3$ $\nearrow$ $\searrow$ Maximum $0$ $\searrow$ $\searrow$ Not an extremum $2$ $\searrow$ $\nearrow$ Minimum Now we can see that $f$ has one relative maximum.